Prove that in any triangle the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median, which bisects the third side.

Ans:    To prove AB² + AC² = 2AD² + 2 (1/2 BC)²

Draw AE ⊥ BC

In Δ ABD since ∠D > 900

∴AB2 = AD² + BD² + 2BD x DE ....(1)

Δ ACD  since ∠ D < 90^o

AC² = AD² + DC² - 2DC x DE   ....(2)

Adding (1) & (2)

AB² + AC² = 2(AD² + BD²)

= 2(AD² + ( 1/2 BC )²)

Or AB² + AC² = 2 (AD² + BD²)

Hence proved